Problem: Graph this system of equations and solve. $-6x+2y = -6$ $2x-2y = -2$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$ $9$ $10$ $\llap{-}2$ $\llap{-}3$ $\llap{-}4$ $\llap{-}5$ $\llap{-}6$ $\llap{-}7$ $\llap{-}8$ $\llap{-}9$ $\llap{-}10$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$ $9$ $10$ $\llap{-}2$ $\llap{-}3$ $\llap{-}4$ $\llap{-}5$ $\llap{-}6$ $\llap{-}7$ $\llap{-}8$ $\llap{-}9$ $\llap{-}10$ Click and drag the points to move the lines.
Explanation: Convert the first equation, $-6x+2y = -6$ , to slope-intercept form. $y = 3 x - 3$ The y-intercept for the first equation is $-3$ , so the first line must pass through the point $(0, -3)$ The slope for the first equation is $3$ . Remember that the slope tells you rise over run. So in this case for every $3$ positions you move up $1$ position to the right. $3$ positions up from $(0, -3)$ is $(1, 0)$ Graph the blue line so it passes through $(0, -3)$ and $(1, 0)$ Convert the second equation, $2x-2y = -2$ , to slope-intercept form. $y = x + 1$ The y-intercept for the second equation is $1$ , so the second line must pass through the point $(0, 1)$ The slope for the second equation is $1$ . Remember that the slope tells you rise over run. So in this case for every $1$ position you move up You must also move $1$ position to the right. $1$ position to the right. $1$ position up from $(0, 1)$ is $(1, 2)$ Graph the green line so it passes through $(0, 1)$ and $(1, 2)$ The solution is the point where the two lines intersect. The lines intersect at $(2, 3)$.